DECOMPOSITION OF (2K + 1)-REGULAR GRAPHS CONTAINING SPECIAL SPANNING 2K-REGULAR CAYLEY GRAPHS INTO PATHS OF
LENGTH 2K + 1
Luiz Henrique Silva Homann
Dissertação de Mestrado apresentada ao Programa de Pós-graduação em Engenharia de Sistemas e Computação, COPPE, da Universidade Federal do Rio de Janeiro, como parte dos requisitos necessários à obtenção do título de Mestre em Engenharia de Sistemas e Computação.
Orientador: Fábio Happ Botler
Rio de Janeiro Fevereiro de 2020
DECOMPOSITION OF (2K + 1)-REGULAR GRAPHS CONTAINING SPECIAL SPANNING 2K-REGULAR CAYLEY GRAPHS INTO PATHS OF
LENGTH 2K + 1
Luiz Henrique Silva Homann
DISSERTAÇÃO SUBMETIDA AO CORPO DOCENTE DO INSTITUTO ALBERTO LUIZ COIMBRA DE PÓS-GRADUAÇÃO E PESQUISA DE ENGENHARIA DA UNIVERSIDADE FEDERAL DO RIO DE JANEIRO COMO PARTE DOS REQUISITOS NECESSÁRIOS PARA A OBTENÇÃO DO GRAU DE MESTRE EM CIÊNCIAS EM ENGENHARIA DE SISTEMAS E COMPUTAÇÃO.
Orientador: Fábio Happ Botler
Aprovada por: Prof. Fábio Happ Botler
Prof. Uéverton dos Santos Souza
Prof. Hugo de Holanda Cunha Nobrega
RIO DE JANEIRO, RJ BRASIL FEVEREIRO DE 2020
Homann, Luiz Henrique Silva
Decomposition of (2k + 1)-regular graphs containing special spanning 2k-regular Cayley Graphs into paths of length 2k + 1/Luiz Henrique Silva Homann. Rio de Janeiro: UFRJ/COPPE, 2020.
VIII, 32 p.: il.; 29, 7cm.
Orientador: Fábio Happ Botler
Dissertação (mestrado) UFRJ/COPPE/Programa de Engenharia de Sistemas e Computação, 2020.
Referências Bibliográ cas: p. 31 32.
1. Decomposition. 2. Path. 3. Regular graph. 4. Cayley graph. I. Botler, Fábio Happ. II. Universidade Federal do Rio de Janeiro, COPPE, Programa de Engenharia de Sistemas e Computação. III. Título.
Resumo da Dissertação apresentada à COPPE/UFRJ como parte dos requisitos necessários para a obtenção do grau de Mestre em Ciências (M.Sc.)
DECOMPOSIÇÃO DE GRAFOS (2K + 1)-REGULARES CONTENDO GRAFOS DE CAYLEY 2K-REGULARES GERADORES ESPECIAIS EM
CAMINHOS DE COMPRIMENTO 2K + 1
Luiz Henrique Silva Homann
Fevereiro/2020
Orientador: Fábio Happ Botler
Programa: Engenharia de Sistemas e Computação
Uma P`-decomposição de um grafo G é um conjunto de caminhos aresta-disjuntas
com ` arestas em G que cobre o conjunto de arestas de G. Favaron, Genest, e Kouider (2010) conjecturaram que todo grafo (2k + 1)-regular que contém um emparelhamento perfeito admite uma P2k+1-decomposição. Eles também veri caram
essa conjectura para grafos 5-regulares sem ciclos de comprimento 4. Em 2015, Botler, Mota, e Wakabayashi estenderam esse resultado para grafos 5-regulares sem triângulos, e em 2017, Botler, Mota, Oshiro e Wakabayashi generalizaram esse resultado para grafos (2k + 1)-regulares com cintura de tamanho pelo menos 2k. Nessa dissertação, veri camos essa conjectura para grafos (2k + 1)-regulares que contêm a k-ésima potência de ciclo; e para grafos 5-regulares que contêm um grafo de Cayley gerador 4-regular gerado por dois elementos comutativos.
Abstract of Dissertation presented to COPPE/UFRJ as a partial ful llment of the requirements for the degree of Master of Science (M.Sc.)
DECOMPOSITION OF (2K + 1)-REGULAR GRAPHS CONTAINING SPECIAL SPANNING 2K-REGULAR CAYLEY GRAPHS INTO PATHS OF
LENGTH 2K + 1
Luiz Henrique Silva Homann
February/2020
Advisor: Fábio Happ Botler
Department: Systems and Computing Engineering
A P`-decomposition of a graph G is a set of edge-disjoint paths with ` edges in G
that cover the edge set of G. Favaron, Genest, and Kouider (2010) conjectured that every simple (2k+1)-regular graph that contains a perfect matching admits a P2k+1
-decomposition. They also veri ed this conjecture for 5-regular graphs without cycles of length 4. In 2015, Botler, Mota, and Wakabayashi extended this result to 5-regular graphs without triangles, and in 2017, Botler, Mota, Oshiro and Wakabayashi generalized this result to (2k + 1)-regular graphs with girth at least 2k. In this dissertation, we verify this conjecture for (2k + 1)-regular graphs that contain the k-th power of the cycle; and for 5-regular graphs that contain spanning 4-regular Cayley graph generated by two commutative elements.
Summary
List of Figures vii
1 Introduction 1
2 Regular graphs that contain spanning powers of cycles 4 3 5-regular graphs that contain spanning Cayley graphs 7 3.1 Preliminary remarks . . . 7 3.2 Complete decompositions . . . 9 3.3 Admissible decompositions . . . 18
4 Conclusion and future works 30
List of Figures
1.1 5-regular graph that admits a P5-decomposition, but has no perfect
matching. . . 2 2.1 P5-decomposition of a 5-regular graph that contains a spanning copy
of a K4,4. . . 4
3.1 The main types of trails. . . 9 3.2 Exchange of edges between elements of type A and a hanging edge in
the proof of Lemma 3. . . 11 3.3 Identities given by Lemma 4 when b3 = 0. . . 11
3.4 Exchange of edges between elements of type A and B in the proof of Claim 1. . . 13 3.5 Exchange of edges between elements of type A and C in the proof of
Claim 1. . . 13 3.6 Exchange of edges between two elements of type A in the proof of
Claim 2. . . 14 3.7 Exchange of edges between three elements of type A in the proof of
Claim 2. . . 15 3.8 Exchange of edges between three elements of type A in the proof of
Claim 2. . . 15 3.9 Exchange of edges between two elements of type A and another
ele-ment of type B in the proof of Claim 2. . . 16 3.10 Exchange of edges between two elements of type A and another
ele-ment of type C in the proof of Claim 2. . . 17 3.11 An exceptional pair. . . 18 3.12 Exchange of edges between the elements an A-chain of type 1 with
ve elements in the proof of Claim 4. . . 20 3.13 Exchange of edges between the elements an A-chain of type 2 with
ve elements in the proof of Claim 4. . . 21 3.14 Exchange performed in the proof of Claim 5 in the case tr(T3) = cv1(T2). 22
3.16 Exchange performed in the proof of Claim 6 in the case tr(T3) =
cv1(T2) . . . 24
3.17 Exchange performed in the proof of Claim 6 in the case tr(T3) =
cv2(T2) . . . 24
3.18 Exchange of edges between two elements of type A and another ele-ment of type E in the proof of Claim 8. . . 26 3.19 Exchange performed in the proof of Claim 6 in the case tr(T3) =
cv1(T2) . . . 27
3.20 Exchange performed in the proof of Claim 6 in the case tr(T3) =
Chapter 1
Introduction
All graphs in this dissertation are simple, i.e., have no loops nor multiple edges. A decomposition of a graph G is a set D of edge-disjoint subgraphs of G that cover its edge set. If every element of D is isomorphic to a xed graph H, then we say that D is an H-decomposition.
The literature related to H-decompositions is quite vast and contains results for decompositions of 2k-regular graphs into trees with k edges. For instance, Ringel [18] conjectured that the complete graph K2`+1 admits a T -decomposition for any tree
with ` edges. Moreover, Ringel's Conjecture holds for many classes of trees (see [6, 11]). Häggkvist [13] generalized Ringel's Conjecture for regular graphs as follows. Conjecture 1 (Graham-Häggkvist, 1989). Let T be a tree with ` edges. If G is a 2`-regular graph, then G admits a T -decomposition.
For more results on decompositions of regular graphs into trees see [7, 9, 14]. For the case T = P`, Kouider and Lonc [16] improved Häggkvist's result proving
that if G is a 2`-regular graph with girth g ≥ (` + 3)/2, then G admits a special P`-decomposition called balanced P`-decomposition, that is a path decomposition D
where each vertex is the end vertex of exactly two paths of D. These authors also stated the following strengthening of Conjecture 1 for paths.
Conjecture 2 (Kouider-Lonc, 1999). Let ` be a positive integer. if G is a 2`-regular graph, then G admits a balanced P`-decomposition.
In this dissertation, we focus on the case in which H is the simple path with 2k + 1 edges, which we denote by P2k+1 (see Figure 2.1). Note that this notation is
not standard. In 1957, Kotzig [15] (see also [3]) proved the following theorem. Theorem 1 (Kotzig, 1957). A 3-regular graph G admits a P3-decomposition if and
Proof. First, suppose that G admits a P3-decomposition D. Since a 3-regular graph
has 3n/2 edges, we have |D| = n/2. Note that every element of D has only one internal edge. Let G0 be the graph consisting of the internal edges of every element
of D. Let P = a0a1a2a3 be a path of D. Since dP(a1) = dP(a2) = 2 and there are
n/2paths in D, dG0(x) = 1 for every x ∈ V (G0). Thus, E(G0)is a perfect matching
of G.
Now, suppose that G has a perfect matching M. Note that, G−M is a 2-regular graph, and hence admits an Eulerian orientation, i.e., an orientation of the edges of G − M in which for each vertex v, precisely half of the edges incident to v are oriented towards v. For each e = xy ∈ M let Pe= a0a1a2a3, where a0a1 and a2a3 are
edges leaving, respectively, a1 and a2. Since G − M admits an Eulerian orientation,
there are no two in-edges of the same vertex. Thus, a0 6= a3 and Pe is a path for
every e ∈ M.
In 2010, Favaron, Genest, and Kouider [8] extended this result by proving that every 5-regular graph that contains a perfect matching and no cycles of length 4 admits a P5-decomposition; and proposed the following conjecture to generalize
Kotzig's result. They also present a 5-regular graph that admits a P5-decomposition
and does not contain a perfect matching (see Figure 1.1). This motivated them to pose the following conjecture.
Conjecture 3 (FavaronGenestKouider, 2010). If G is a (2k + 1)-regular graph that contains a perfect matching, then G admits a P2k+1-decomposition.
Figure 1.1: 5-regular graph that admits a P5-decomposition, but has no perfect
matching.
In 2015, Botler, Mota, and Wakabayashi [2] veri ed Conjecture 3 for triangle-free 5-regular graphs, and, more recently, Botler, Mota, Oshiro, and Wakabayashi [1] generalized this result for (2k + 1)-regular graphs with girth at least 2k.
It is clear that a 5-regular graph contains a perfect matching if and only if it contains a spanning 4-regular graph. In fact, by using a theorem of Petersen [17], one can prove that a (2k + 1)-regular graph contains a perfect matching if and only if it contains a spanning 2k0-regular graph for every k0 ≤ k.
Theorem 2 (Petersen, 1891). If G is a 2k-regular graph, then G admits a decom-position into spanning 2-regular graphs.
In this dissertation, we explore Conjecture 3 for (2k + 1)-regular graphs that contain special spanning 2k-regular graphs as follows. Throughout the text, Γ de-notes a nite group of order n; + dede-notes the group operation of Γ; and 0 dede-notes the identity of Γ. As usual, for each x ∈ Γ, we denote by −x the element y ∈ Γ such that x + y = 0, and the operation − denotes the default binary operation − : (x, y) 7→ x + (−y). Let S ⊆ Γ be a set not containing the identity of Γ, and such that −x ∈ S for every x ∈ S (i.e., S is closed under taking inverses). The Cayley graph X(Γ, S) is the graph H with V (H) = Γ, and E(H) = {x, y}: y − x ∈ S (see [10]). 1 In this dissertation, we consider the general case in which S is not a
set generating Γ, and hence X(Γ, S) is not necessarily connected. We say that H is simply commutative if (i) x+y = y +x for every x, y ∈ S, and (ii) x+y 6= 0 for every x, y ∈ S with y 6= −x. Since 0 /∈ S, condition (ii) guarantees that H is a simple graph. It is not hard to check that, in such a graph, the neighborhood of a vertex v ∈ Γ is N(v) = S + v = {x + v : x ∈ S}. Although the de nition of Cayley graphs can be extended to multigraphs and directed graphs, in this dissertation we consider only non-directed simple graphs in order to tackle Conjecture 3. The reader may regard the graphs studied here as simple graphs that contain underlying graphs of directed Cayley graphs.
In this dissertation, we present two results regarding Conjecture 3. We verify it for (2k + 1)-regular graphs that contain the k-th power of a cycle (see Chapter 2); and for 5-regular graphs that contain spanning simply commutative 4-regular Cayley graphs (see Chapter 3).
We believe that, due to the underlying group structure, the techniques developed here can be extended for dealing with (2k + 1)-regular graphs that contain more general spanning Cayley graphs, and also (2k+1)-regular graphs that contain special spanning Schreier Graphs, which can give us signi cant insight with respect to the general case of Conjecture 3 (see Chapter 4).
1 Although this de nition is well-known, there is another de nition which give us a colored
directed graph as Cayley Graph, let X be a group and S a generating set of X, the Cayley Graph
Γ = Γ(X, S)is constructed as follows: (i) X = V (Γ); (ii) each generator s of S is assigned a color
cs; (iii) for any x ∈ X and s ∈ S, the vertices corresponding to the elements x and xs are joined by
a directed edge of color cs. Thus, the edge set E(Γ) consists of pairs of the form (x, xs) in which
Chapter 2
Regular graphs that contain
spanning powers of cycles
Given positive integers k and n, the k-th power of the cycle on n vertices, which we denote by Ck
n, is the graph on the vertex set {0, . . . , n − 1} and such
that, for every vertex v, we have x ∈ N(v) if and only if x = v + r (mod n), where r ∈ {−k, . . . , −1} ∪ {1, . . . , k}. Given a perfect matching M of a graph G we say that a P`-decomposition D of a graph G is M-centered if for every
P = a0a1. . . aiai+1. . . a`−1a` ∈ D, we have aiai+1 ∈ M for i = (` − 1)/2. The
next results are examples of M-centered decomposition.
Proposition 1. If G is a 5-regular graph that contains a spanning copy K of K4,4,
and M = G − E(K), then G admits a M-centered P5-decomposition.
Proof. Let G, K, and M be as in the statement. Let (R, L) be the bipartition of K, where R = {r1, r2, r3, r4} and L = {l1, l2, l3, l4} be the partition classes of K.
Since K is a complete bipartite graph, if xy ∈ M, then either x, y ∈ R or x, y ∈ L. Thus, we may suppose, without loss of generality, that M = {r1r2, r3r4, l1l2, l3l4}.
Therefore, D = {l1r1l3l4r2l2, l3r3l1l2r4l4, r1l2r3r4l1r2, r3l4r1r2l3r4}is an M-centered
decomposition of G as desired (see Figure 2.1).
r3 r2 r4 r1 l3 l2 l4 l1
Figure 2.1: P5-decomposition of a 5-regular graph that contains a spanning copy of
Proposition 2. Let n, k ∈ N with k < n/2. If G is a simple (2k + 1)-regular graph on n vertices that contains a copy C of Ck
n, and M = G − E(C), then G admits an
M-centered P2k+1-decomposition.
Proof. Let G, C, and M be as in the statement, and let V (C) = {0, . . . , n − 1} as above. Since C is a 2k-regular graph, M is a perfect matching of G. Given i ∈ V (C), let Qi be the path v0v1. . . vk in which v0 = i; and, for j = 1, . . . , k, we
have vj = vj−1+ j if j is odd; and vj = vj−1− j if j is even. Note that for every
j = 1, . . . , k, the path Qi contains an edge xy such that |x − y| = j. Also, we have
V (Qi) =i + r (mod n) : r ∈ {−bk/2c, −bk/2c + 1, . . . , dk/2e} . It is not hard to
check that the set Q = Qi: i ∈ V (C)
is a Pk-decomposition of C.
Given an edge e = ij ∈ M, let Pe = Qi ∪ {ij} ∪ Qj. Since Qi and Qj have,
respectively, i and j as end vertices, and E(Qi) ∩ E(Qj) = ∅, the graph Pe is a
trail of length 2k + 1. Thus, since Q is a Pk-decomposition of C, and M is a perfect
matching of G, the set D = {Pe: e ∈ M }is a decomposition of G into trails of length
2k + 1. We claim that, in fact, D is a P2k+1-decomposition of G. For that, we prove
that if ij ∈ M, then V (Qi) ∩ V (Qj) = ∅. Indeed, note that for every e = ij ∈ M, we
have |i − j| > k, otherwise we have ij ∈ E(C). Now, suppose that there is a vertex v in V (Qi) ∩ V (Qj). Then, there are r1, r2 with −bk/2c ≤ r1, r2 ≤ dk/2e, and such
that i + r1 = v = j + r2. Suppose, without loss of generality, that i > j. Then, we
have r2− r1 = i − j > k, but r2− r1 ≤ dk/2e + bk/2c = k, a contradiction.
Note that, from the proof of Proposition 2 we also obtain a construction for the Hamilton path decomposition of complete graphs of even order.
Corollary 1. The complete graph K2k+2 admits a P2k+1-decomposition.
A slight variation of the proof of Proposition 2 also provides the following result. Corollary 2. Let G be an `-regular graph, ` odd, and let M be a perfect matching of G. If each component of G \ M is a k-th power of a cycle, then G admits an M-centered P`-decomposition.
Let G1 and G2 be disjoint graphs with perfect matchings M1 and M2,
re-spectively. Let a1b1, . . . , akbk ∈ M1 and x1y1, . . . , xkyk ∈ M2 be distinct edges,
and let G be the graph obtained from the disjoint union G1 ∪ G2 by
remov-ing a1b1, . . . , akbk, x1y1, . . . , xkyk and adding the edges a1x1, b1y1, . . . , akxk, bkyk.
We say that G is a collage of G1 and G2 over edges of M1 and M2, and
denote by MG the perfect matching M1 ∪ M2 ∪ {a1x1, b1y1, . . . , akxk, bkyk} \
{a1b1, . . . , akbk, x1y1, . . . , xkyk}. When M1 and M2 are clear from the context, we
say simply that G is a collage of G1 and G2. Note that G is `-regular if and only if
Let G be an `-regular graph, where ` is an odd positive integer, and let M be a perfect matching of G. We say that G is M-constructable if either G admits an M-centered P`-decomposition, or G is the collage of an M1-constructable graph and
an M2-constructable graph over edges of M1 and M2 and M = MG. The next result
is a useful tool in the proof of Theorem 3.
Lemma 1. Every M-constructable `-regular graph admits an M-centered P`
-decomposition.
Proof. Suppose, for a contradiction, that the statement does not hold, and let G be a counterexample with a minimum number of vertices. By the de nition of M-constructable, G is the collage of an M1-constructable graph G1 and an
M2-constructable graph G2 over edges of M1 and M2. By the minimality of
G, the graph Gi admits an Mi-centered P`-decomposition Di, for i = 1, 2. Let
ai, bi, xi, yi, for i = 1, . . . , k be such that G is the graph obtained from G1 ∪ G2
by removing a1b1, . . . , akbk, x1y1, . . . , xkyk and adding a1x1, b1y1, . . . , akxk, bkyk as
above. For i = 1, . . . , k, let Pi ∈ D1 and Qi ∈ D2 be the paths
contain-ing the edges aibi and xiyi, respectively. By the de nition of M1- and M2
-centered P`-decomposition, for i = 1, . . . , k, we may write Pi = Pi,1aibiPi,2 and
Qi = Qi,1xiyiQi,2, where Pi,1, Pi,2, Qi,1 and Qi,2 are paths of length (` − 1)/2. Since
G1 and G2 are disjoint, V (Pi,j) ∩ V (Qi,j) = ∅ for i = 1, . . . , k and j = 1, 2.
Let Ri,1 = Pi,1 ∪ {aixi} ∪ Qi,1 and Ri,2 = Pi,2 ∪ {biyi} ∪ Qi,2, and note that
D = D1\ {P1, . . . , Pk} ∪ D2\ {Q1, . . . , Qk} ∪ {Ri,j: i = 1, . . . , k and j = 1, 2} is
an MG-centered P`-decomposition of G, a contradiction.
Corollary 3. If G is a 5-regular graph that contains a K4,4-factorization K and
Chapter 3
5
-regular graphs that contain
spanning Cayley graphs
In this chapter, we explore 5-regular graphs that contain spanning simply com-mutative 4-regular Cayley graphs. In [2], Botler et al. showed that every triangle-free 5-regular graph G that has a perfect matching admits a P5-decomposition. For
that, they applied the following strategy: i) to nd an initial decomposition of G into paths and trails; and ii) to perform some exchange of edges on the elements of D preserving a special invariant, while minimizing the number of trails that are not paths. The proof of our main theorem follows this framework, but consists of three steps. First, from the structure of Cayley graphs, we nd an initial decompo-sition into trails, not necessarily paths (see Propodecompo-sition 3). Then, we show how to exchange edges to obtain a decomposition in which the bad elements (the trails that are not paths) are distributed in circular fashion (see Lemma 5). Finally, we show how to deal with these cycles of bad elements (see Theorem 3).
3.1 Preliminary remarks
In this section, we present four special types of trails used throughout this text, and the following useful lemma.
Lemma 2. Let k ∈ N. If G is a (2k + 1)-regular graph, and D is a decomposition of G into trails of length 2k + 1, then each vertex of G is the end vertex of precisely one element of D.
Proof. Let k, G and D be as in the statement. Let n = |V (G)|. Given an element T ∈ D, we denote by o(T ) the number of its vertices with odd degree in T , and given a vertex v ∈ V (G), we denote by D(v) the number of trails in D having v as end vertex. Clearly, PT ∈Do(T ) =
P
v∈V (G)D(v). Since T is a trail, we have
|D| = 1 2k+1|E(G)| = 1 2k+1 1 2(2k + 1)n = 1
2n. Thus, we have PT ∈Do(T ) ≤ 2|D| = n.
Now, since v ∈ V (G) has odd degree, v must have odd degree in at least one element of D, and hence D(v) ≥ 1. Thus, we have Pv∈V (G)D(v) ≥ n, and hence
P
v∈V (G) =
P
T ∈Do(T ) = n. This implies that D(v) = 1 for every v ∈ V (G), as
desired.
Recall that Γ is a nite group of order n with operation +. Given two ele-ments g, r of Γ, we say that {g, r} is a simple commutative generator (SCG) if (i) 0 /∈ {g, r, 2g, 2r}; (ii) g /∈ {r, −r}; and (iii) g + r = r + g. Let S = {g, −g, r, −r}, and consider the Cayley graph C = X(Γ, S). By construction, C is a simply com-mutative Cayley graph (see Chapter 1). Conditions (i) and (ii) guarantee that C is a simple graph, while condition (iii) introduces the main restriction explored in this dissertation. In this case, we say that C is the graph generated by {g, r}, and that {g, r} is the generator of C. Given an SCG {g, r}, we say that a simple 5-regular graph G with vertex set Γ is a {g, r}-graph if G contains a spanning Cayley graph C generated by {g, r}. We say that G is a simply commutative generated graph or, for short, SCG-graph if G is a {g, r}-graph for some SCG {g, r}. In this chapter, we verify Conjecture 3 for SCG-graphs.
A 2-factor in a graph G is a spanning 2-regular subgraph of G. Let {g, r} be an SCG. If C is the graph generated by {g, r}, and x ∈ {g, r}, then we denote by Fx
the 2-factor of C with edge set E(Fx) = {v + x : v ∈ Γ}. If G is a {g, r}-graph, then
we denote by Mg,r the perfect matching G\E(Fg∪ Fr), and the triple {Mg,r, Fg, Fr}
is called the base factorization of G. Although G is a simple graph, for ease of notation, we refer to an edge uv ∈ Fx, with x ∈ {g, r}, as a green (resp. red)
out-edge of u and in-edge of v if v = u + x and x = g (resp. x = r). In the gures throughout the text, the edges in Fg, Fr, Mg,r are illustrated, respectively, in green,
red, and double black pattern, while edges without speci c a liation are illustrated in gray. Moreover, if such an edge has a speci c direction (i.e., in-edge or out-edge), it is illustrated accordingly. Note that each vertex u has precisely one edge of each type (green in-edge, green out-edge, red in-edge, red out-edge), and is incident to precisely one edge of Mg,r. Note that the group structure overcomes Theorem 2 by
giving a decomposition of C into 2-factors in terms of the elements g and r. The next de nition presents the main elements of the decompositions in our proofs. De nition 1. We say that a trail T in a {g, r}-graph is of type A, B, C, or D if T can be written as a0a1a2a3a4a5 as follows.
type A: a0, a1, a2, a3, a4 are distinct vertices and a2 = a5 and the following hold:
a2a3 ∈ Mg,r, a2a1, a3a4 ∈ Fg, a5 = a4a5, and a1a0 ∈ Fg∪Fr∪Mg,r, i.e., a1a0
is an out-edge of a1, or a1a0 ∈ Mg,r (see Figure 3.1a). In this case, we say
vertex of T ; a1 is the auxiliary vertex of T ; and a4 is the tricky vertex of
T. We denote these vertices, respectively, by cv1(T ), cv2(T ), aux(T ), and
tr(T );
type B: a0, a1, a2, a3, a4, a5 are distinct vertices and the following hold: a2a3 ∈ Mg,r
a2a1, a3a4 ∈ Fg, a1a0, a4a5 ∈ Fg∪ Fr∪ Mg,r (see Figure 3.1b);
type C: a0, a1, a2, a3, a4, a5 are distinct vertices and the following hold: a2a1, a4a3 ∈
Fg, a3a2, a4a5 ∈ Fr, a1a0 ∈ Fg∪ Fr∪ Mg,r, and, moreover, we have a2a4 ∈
E(G) and a2a4 ∈ Mg,r (see Figure 3.1c);
type D: a0, a1, a2, a3, a4, a5 are distinct vertices and the following hold: a1a0, a4a5 ∈
Fr, a1a2, a3a4 ∈ Mg,r, and a3a2 ∈ Fg (see Figure 3.1d).
a0 a1 a2= a5 a3 a4 a4 a5 (a) Type A. a0 a1 a2 a3 a4 a5 (b) Type B. a0 a1 a2 a4 a3 a5 (c) Type C. a0 a1 a4 a3 a2 a5 (d) Type D.
Figure 3.1: The main types of trails.
We remark that elements of type A are not paths, while elements of type B, C, and D are paths. Moreover, the connection vertices of an element T are always incident to an edge of Mg,rin T , and hence, no vertex of a {g, r}-graph is a connection
vertex of two edge-disjoint elements of type A in a graph.
Given a trail (not necessarily a path) T = a0a1a2a3a4a5 in a decomposition D
of a {g, r}-graph G, we say that the edge a1a0 (resp. a4a5) is a hanging edge at a1
(resp. a4) if a1a0 ∈ Mg,r∪Fg∪Fr(resp. a4a5 ∈ Mg,r∪Fg∪Fr), i.e., the hanging edges
of T are the end edges of T that are in Mg,r or that are in-edges of its end vertices.
By De nition 1, all end edges of elements of type A, B, C, or D are hanging edges. Note that if T is an element of type A where a5 = a2, then a1a0, a2a3 and a4a2 are
hanging edges of T at, respectively, a1, a3, and a4. Given a trail decomposition D
of a graph G and a vertex u ∈ V (G), we denote by hangD(u) the number of edges
of G that are hanging edges at u.
3.2 Complete decompositions
The proof of the main theorem of this dissertation relies on the following de ni-tion, which consist of a set of properties that hold for decompositions of the given
graph, and that are invariant due to a series of operations performed throughout the proof.
De nition 2. A decomposition D of a {g, r}-graph G into trails of length 5 is complete if the following hold.
(a) hangD(u) > 0 for every vertex u that is a connection vertex of an element of D;
(b) Every element in D is of type A, B, C or D;
(c) If T can be written as a0a1a2a3a4a5 and a0a1 ∈ Mg,r, then a0 = a1+ g + r.
The rst step of our proof is given by the next proposition, which presents an initial decomposition for the graphs studied.
Proposition 3. If G is an SCG-graph, then G admits a complete decomposition. Proof. Let {Fg, Fr, Mg,r} be the base factorization of G. For each e = xy ∈ Mg,r,
let Pe = a0a1a2a3a4a5, where a1a0, a4a5 ∈ Fr, a2a1, a3a4 ∈ Fg, a2 = x, and a3 = y.
We claim that D = {Pe: e ∈ Mg,r} is complete. Clearly, Pe is an element of type A
or B, for every e ∈ Mg,r, and hence D satis es De nition 2(b). Moreover, note that
a0a1 (resp. a4a5) is a hanging edge at a1 (resp. a4). Thus, given z ∈ V (G), let e0 =
xy ∈ Mg,r be such that x = z − g, then Pe0 contains a hanging edge at z, and hence
there is a hanging edge at every vertex of G. This proves De nition 2(a). Finally, if P ∈ D can be written as a0a1a2a3a4a5 where a0a1 ∈ Mg,r, then P must be of type
A, and we have a1a2, a3a4 ∈ Fg, a2a3, a4a5 ∈ Fr and a0 = a3 = a2+ r = a1+ g + r,
and hence D satis es De nition 2(c).
The next lemma is used often in our proof and presents a consequence of the exchange of hanging edges at primary connection vertices.
Lemma 3. If T = a0a1a2a3a4a5 is an element of type A in a decomposition of a
{g, r}-graph G into trails of length 5, where a5 = a2 and a2a3 ∈ Mg,r, and u ∈ V (G)
is such that a3u is a hanging edge at a3 = cv1(T ), then T0 = a0a1a2a4a3u is of type
C.
Proof. Let T , u, and T0 be as in the statement. Since a
3a4 is a green out-edge of a3
and a2a3 is an edge of Mg,r incident to a3, we conclude that a3u is a red out-edge
of a3, and hence u = a3 + r. Now, since G is simple, we have u /∈ {a2, a3, a4}; if
u = a1, then we have a3 + r = u = a1 = a3 + g + r + g, which implies 2g = 0, a
contradiction to the de nition of SCG. Finally, by Lemma 2 we have u 6= a0. Thus,
a0 a1 a2 a3 a4 u a0 a1 a2 a3 a4 u
Figure 3.2: Exchange of edges between elements of type A and a hanging edge in the proof of Lemma 3.
The following lemma is also used often in our proof and shows how two elements of type A may be connected.
Lemma 4. If T1 and T2 are two edge-disjoint elements of type A in a {g, r}-graph
G such that tr(T2) = cv1(T1), then aux(T2) = cv2(T1).
Proof. Let T1 = a0a1a2a3a4a5 and T2 = b0b1b2b3b4b5, where a5 = a2 and b5 = b2
and a2a3, b2b3 ∈ Mg,r. If cv1(T1) = tr(T2), then a3 = b4. Since b1 = b4+ r + g and
a2 = a3+g+r. Thus aux(T2) = b1 = b4+r+g = a3+r+g = a3+g+r = a2 = cv2(T1),
as desired (see Figure 3.3).
g 2g
2g + r
g + r 0
Figure 3.3: Identities given by Lemma 4 when b3 = 0.
We say that an element T of type A in a complete decomposition D is free if tr(T ) 6= cvi(T0) for every T0 ∈ D and i ∈ {1, 2}. An A-chain is a sequence
T0, T1, . . . , Ts−1 of elements of type A such that for each j ∈ {0, . . . , s − 1}, we
have tr(Tj) = cvi(Tj−1), for some i ∈ {1, 2} (subtraction on the indices are taken
modulo s). Note that A-chains do not consider the auxiliary vertex when allowing two elements to be consecutive. Thus, elements, say T1 and T2, of type A that are
not consecutive in an A-chain, or that are in dierent A-chains, may still share a vertex u for which cvi(T1) = u = aux(T2).
Given a decomposition D of a graph G into trails of length 5, denote by τ(D) the number of elements that are not paths. By exchanging edges between the elements of a decomposition given by Proposition 3, we can show that a complete decomposition that minimizes τ(D) has no free element, and hence its elements of type A are partitioned into A-chains.
Lemma 5. Every {g, r}-graph for which 2g + 2r 6= 0 admits a complete decomposi-tion in which the elements of type A are partidecomposi-tioned into A-chains.
Proof. Let G be a {g, r}-graph for which 2g +2r 6= 0 and put M = Mg,r. By
Propo-sition 3, G admits a complete decompoPropo-sition. Let D be a complete decompoPropo-sition of Gthat minimizes τ(D). In what follows, we prove that D contains no free element of type A.
For that, in each step, we exchange edges between some elements of D and obtain a complete decomposition D0 into trails of length ve such that τ(D0) < τ (D),
which is a contradiction to the minimality of D. To check that D0 is a complete
decomposition, we observe the three following items: (i) The only vertex that has a hanging edge of D but does not have a hanging edge of D0 is the tricky vertex of a
free element of type A, which by the de nition of free element, is not a connection vertex of any element of D0, and hence De nition 2(a) holds for D0; (ii) every element
of D0 that is not an element of D, i.e., the elements involved in the exchange of edges,
is of type A, B, C, or D, and hence 2(b) holds for D0; (iii) every edge uv ∈ M in
the elements considered that can be viewed as an end edge in D0 (some elements
may be expressed by more than two trails) either can be viewed as end edges in D or is obtained from an element of type A and satis es u = v + g + r, and hence De nition 2(c) holds for D0.
Claim 1. No element of type B or C has a hanging edge at the primary connection vertex of a free element of type A.
Proof. Let T1 ∈ D be a free element of type A, and let T2 ∈ D be an element of
type B or C that contains a hanging edge at cv1(T1). Let T1 = a0a1a2a3a4a5, where
a5 = a2 and a2a3 ∈ M, and T2 = b0b1b2b3b4b5. We divide the proof depending on
whether T2 is of type B or C.
T2 is of type B. Suppose, for a contradiction, that b4 = cv1(T1) = a3. Put
T10 = a0a1a2a4a3b5, T20 = b0b1b2b3b4a2 (see Figure 3.4), and let D0 = D \ {T1, T2} ∪
{T0 1, T
0
2}. Note that D
0 is a decomposition of G into trails of length 5. By Lemma 3,
T10 is an element of Type C. In what follows, we prove that T20 is a path. For that, we prove that a2 ∈ {b/ 0, b1, b2, b3, b4}. Indeed, since G has no loops or multiple edges,
a2 ∈ {b/ 3, b4}. Since M is a matching, a2 6= b2. If a2 = b1, then b2 = b1− g = a2− g =
a3+ g + r − g = b5, and hence T2 is of type A, a contradiction. Finally, by Lemma 2
a2 6= b0. Thus, T20 is a path. Moreover, T 0
2 is an element of type B.
In what follows, we prove that D0 is a complete decomposition. Note that
hangD0(v) ≥ hangD(v) for every v ∈ V (G) \ {a4}. Since a4 is not a connection
vertex of D0, and hang
D(v) > 0 for every connection vertex v of D, hangD0(v) > 0
for every connection vertex v of D0. Thus De nition 2(a) holds for D0. Also, since
T10 is of type C and T20 is of type B, De nition 2(b) holds for D0. Moreover, note that a2a3 ∈ M is an end edge of T20. Since T1 is of type A, we have a2 = a3+ g + r.
that τ(D0) = τ (D) − 1 < τ (D), a contradiction to the minimality of D. a0 a1 a2 a3 a4 b0 b1 b2 b3 b4 b5 a0 a1 a2 a3 a4 b0 b1 b2 b3 b4 b5
Figure 3.4: Exchange of edges between elements of type A and B in the proof of Claim 1.
T2 is of type C. We may assume b3b2 ∈ Fr. In this case we have b4b3 ∈ Fg.
Since T2 contains a hanging edge at cv1(T1), we have a3 = cv1(T1) ∈ {b1, b4}. If
b4 = a3, then there are two green out-edges at a3, namely a3a4, b4a3, a contradiction.
Thus, we may assume that a3 = b1. Put T10 = a0a1a2a4a3b0, T20 = a2b1b2b4b3b5 (see
Figure 3.5), and let D0 = D \ {T
1, T2} ∪ {T10, T 0
2}. Note that D
0 is a decomposition
of G into trails of length 5. By Lemma 3, T0
1 is an element of Type C. In what
follows we prove that T0
2 is a path. For that, we prove that a2 ∈ {b/ 0, b1, b2, b3, b4}.
Indeed, since G is simple, a2 ∈ {b/ 1, b2}. If a2 = b4, then a2a1 and b4b3 are two green
out-edges at a2, a contradiction. By Lemma 2, a2 6= b5. Finally, a3+ g + r = a2 and
b1 = b3+ r + g, if a2 = b3, then we have a3+ g + r = a2 = b3 = b1− g − r = a3− g − r,
which implies 2g + 2r = 0, a contradiction. Thus T0
2 is a path. Moreover, T20 is an
element of type C.
Now, we prove that D0 is a complete decomposition. Analogously to the case
above hangD0(v) ≥ hangD(v)for every v ∈ V (G)\{a4}. Since a4 is not a connection
vertex of D0, and hang
D(v) > 0 for every connection vertex v of D, hangD0(v) > 0
for every connection vertex v of D0. Thus De nition 2(a) holds for D0. Also, since
T10 and T20 are both elements of type C, De nition 2(b) holds. Moreover, note that a2a3 is an end edge of T20, and, by de nition of element of type A, De nition 2(c)
holds. Thus, D0 is a complete decomposition such that τ(D0) = τ (D) − 1 < τ (D), a
contradiction to the minimality of D.
a0 a1 a2 a3 a4 b5 b4 b3 b2 b1 b0 a0 a1 a2 a3 a4 b5 b4 b3 b2 b1 b0
Figure 3.5: Exchange of edges between elements of type A and C in the proof of Claim 1.
Claim 2. Let T1 and T2 be two elements of type A in D. If T1 is free and T2 contains
a hanging edge on cv1(T1), then no element of type A, B, or C in D\{T1, T2}contains
a hanging edge at cv2(T2).
Proof. Let T1 = a0a1a2a3a4a5 and T2 = b0b1b2b3b4b5 be two elements of D, where
a5 = a2 and b5 = b2 and a2a3, b2b3 ∈ M. First, we prove that cv1(T1) = tr(T2) and
cv2(T1) = aux(T2). Suppose, for contradiction, that cv1(T1) 6= tr(T2). Therefore,
b1 = cv1(T1) = a3. Now, put T10 = a0a1a2a4a3b0, T20 = a2b1b2b3b4b2 (see Figure 3.6)
and let D0 = D \ {T
1, T2} ∪ {T10, T 0
2}. Note that D
0 is a decomposition of G into
trails of length 5. By Lemma 3, T0
1 is an element of type C. We claim that T 0 2 is an
element of type A. For that we prove that a2 ∈ {b/ 1, b2, b3, b4}. Again, since G is a
simple graph, we have a2 ∈ {b/ 1, b2}. Since every vertex is incident to one edge of
M, we have a2 6= b3, and if a2 = b4, then we have a3+ g + r = a2 = b4 = a3− g − r,
which implies 2g + 2r = 0, a contradiction. Now, we prove that D0 is a complete
decomposition. Analogously to the cases above, we have hangD0(v) ≥ hangD(v) ≥ 0
for every v ∈ V (G) \ {a4}. Therefore, De nition 2(a) holds for D0. Moreover,
a3b0 ∈ M/ and if a0a1 ∈ M, then, by De nition 2(a), we have a0 = a1 + g + r,
and hence De nition 2(c) holds for D0. Thus, D0 is a complete decomposition such
that τ(D0) = τ (D) − 1 < τ (D), a contradiction to the minimality of D. Finally, by
Lemma 4, we have cv2(T1) = aux(T2).
a0 a1 a2 a3= b1 a4 b0 b2 b3 b4 a0 a1 a2 a3= b1 a4 b0 b2 b3 b4
Figure 3.6: Exchange of edges between two elements of type A in the proof of Claim 2.
Now, let T3 ∈ D \ {T1, T2} be an element of type A, B, or C, and suppose, for
a contradiction, that T3 contains a hanging edge at cv2(T2). Since cv1(T1) = tr(T2)
and cv2(T1) = aux(T2), we have a3 = b4, a5 = a2 = b1 and b5 = b2. In what follows,
we divide the proof according to the type of T3.
T3 is of type A. Let T3 = c0c1c2c3c4c5, where c2 = c5 and c2c3 ∈ M. Since each
vertex is incident to precisely one edge of M we have c3 6= b2 = cv2(T2). Therefore
we have cv2(T2) ∈ {c1, c4}. Suppose that cv2(T2) = c4. Thus, we have b5 = b2 = c4.
Put T0
1 = a0a1a2a4a3b5, T20 = b0b1b4b3b2c2, T30 = b1c4c3c2c1c0 (see Figure 3.7), and
let D0 = D \ {T 1, T2, T3} ∪ {T10, T 0 2, T 0 3}. Note that D 0 is a decomposition of G into
trails of length 5. In what follows, we prove that T0 1, T
0
2 and T 0
3 are all paths. By
Lemma 3, T0
and b1 ∈ {c/ 2, c3, c4}. By Lemma 2, b2 6= a0, c2 6= b0, b1 6= c0. Therefore, T20 is an
element of type D. If b1 = c1, then b2b1 and c2c1 are two green in-edges at c1, a
contradiction. Therefore, T0
3 is an element of type B.
We claim that D0 is a complete decomposition. Analogously to the cases above,
we have hangD0(v) ≥ hangD(v) ≥ 0 for every v ∈ V (G) \ {a4}. Therefore,
De ni-tion 2(a) holds for D0. Also, since T0 1, T
0 2 and T
0
3 are of type C, D and B, respectively,
De nition 2(b) holds for D0. Moreover, none of the edges in E(T
1) ∪ E(T2) ∪ E(T3)
that may be in M and is not an end edge of T1, T2, or T3, namely a0a1 and c0c1,
became an end edge of T0
1, T20, or T30. Thus De nition 2(c) holds for D0. Thus, D0 is
a complete decomposition such that τ(D0) = τ (D) − 1 < τ (D), a contradiction to
the minimality of D. a0 a1 a3= b4 a4 b0= c1 b1= a2 b2= c4 b3 c0 c2 c3 a0 a1 a3= b4 a4 b0= c1 b1= a2 b2= c4 b3 c0 c2 c3
Figure 3.7: Exchange of edges between three elements of type A in the proof of Claim 2.
Thus, we may assume cv2(T2) = c1. This implies that b5 = b2 = c1, and hence
we have b3 = b5− r − g = c1− g − r = c4. Put T10 = a0a1a2a4a3b2, T20 = b0b1b4b3b2c0,
T30 = b1c1c2c3c4c2 (see Figure 3.8) and D0 = D \ {T1, T2, T3} ∪ {T10, T20, T30}. Note
that D0 is a decomposition of G into trails of length 5. Again, by Lemma 3, T0 1 is an
element of type C. We claim that T0
2, T30 are, respectively, of type D and A. Since
Gis simple, c0 ∈ {b/ 1, b2, b3, b4} and b1 ∈ {c/ 1, c2, c4}. By Lemma 2 we have c0 6= b0.
Therefore, T0
2 is of type D. Finally, if b1 = c3, then a2a1 and c3c4 are two green
out-edges at c3, a contradiction. Therefore, T30 is an element of type A. Analogously to
the case above, D0 is a complete decomposition such that τ(D0) = τ (D) − 2 < τ (D),
a contradiction to the minimality of D.
a0 a1 a3= b4 a4 b0 b1= a2 b3= c4 c0 c1= b2 c2 c3 a0 a1 a3= b4 a4 b0 b1= a2 b3= c4 c0 c1= b2 c2 c3
Figure 3.8: Exchange of edges between three elements of type A in the proof of Claim 2.
T3 is of type B. Let T3 = c0c1c2c3c4c5 be an element of type B. Since T3 contains
a hanging edge on cv2(T2) = b2, we have b2 ∈ {c1, c4}. By symmetry we may
assume b2 = c1. Thus, put T10 = a0a1a2a4a3b2, T20 = b0b1b4b3b2c0, T30 = b1c1c2c3c4c5
(see Figure 3.9) and let D0 = D \ {T
1, T2, T3} ∪ {T10, T 0 2, T 0 3}. Note that D 0 is a
decomposition of G into trails of length 5. Again, by Lemma 3, T0
1 is an element
of type C. We prove that T0
2 and T 0
3 are all paths. Since G is simple, we have
c0 ∈ {b/ 1, b2, b3, b4} and b1 ∈ {c/ 1, c2}. By Lemma 2, we have c0 6= b0 and b1 6= c5.
Therefore, T0
2 is an element of type D. Since c4 = c3+ g and b1 = c1+ g, if c4 = b1,
then c3 = c1, a contradiction. If b1 = c3, then a2a1 and c3c4 are two green out-edges
of b1, a contradiction, and if b1 = c4, then c1b1 and c3c4 are two green in-edges of b1,
again a contradiction. Therefore, T0
3 is an element of type B. Analogously to the
case above, D0 is a complete decomposition such that τ(D0) = τ (D) − 2 < τ (D), a
contradiction to the minimality of D.
a0 a1 a3= b4 a4 b0 b1= a2 b3 c0 c1= b2 c2 c3 c4 c5 a0 a1 a3= b4 a4 b0 b1= a2 b3 c0 c1= b2 c2 c3 c4 c5
Figure 3.9: Exchange of edges between two elements of type A and another element of type B in the proof of Claim 2.
T3 is of type C. Let T3 = c0c1c2c3c4c5 be an element of type C, where c3c2 ∈ Fr.
This implies that c4c3 ∈ Fg. Since T3 contains a hanging edge on cv2(T2) = b2,
we have b2 ∈ {c1, c4}. If b2 = c4, then c4c3 and b2b1 are two green out-edges
of b2, a contradiction. Thus, we may assume b2 = c1. Put T10 = a0a1a2a4a3b2,
T20 = b0b1b4b3b2c0, T30 = b1c1c2c3c4c5(see Figure 3.10) and let D0 = D\{T1, T2, T3}∪
{T0 1, T 0 2, T 0 3}. Note that D
0 is a decomposition of G into trails of length 5. Again, by
Lemma 3, T0
1 is an element of type C. We prove that T20 and T30 are all paths. Since
G is simple, c0 ∈ {b/ 1, b2, b3, b4} and b1 ∈ {c/ 1, c2}. By Lemma 2, we have c0 6= b0
and b1 6= c5. Therefore, T20 is an element of type D. Analogously to the case above,
if b1 = c3, then b2b1 and c4c3 are two green in-edges of b1, a contradiction, and if
b1 = c4, then a2a1 and c4c3 are two green out-edges of b1, again a contradiction.
Therefore, T0
3 is an element of type C. Once more, analogously to the cases above,
D0 is a complete decomposition such that τ(D0) = τ (D) − 2 < τ (D), a contradiction
a0 a1 a3= b4 a4 b0 b1= a2 b3 c0 c1= b2 c2 c3 c4 c5 a0 a1 a3= b4 a4 b0 b1= a2 b3 c0 c1= b2 c2 c3 c4 c5
Figure 3.10: Exchange of edges between two elements of type A and another element of type C in the proof of Claim 2.
Claim 3. There is no free element of type A.
Proof. Suppose, for a contradiction, that D contains a free element, say T1, of type
A. By De nition 2(a), there is a hanging edge e2 on cv1(T1). Let T2 be the element
of D that contains e2. By Claim 1, T2 is not of type B or C, and since M is a
matching, T2 is not of type D. Thus, T2 is of type A. By De nition 2(a), there
is a hanging edge e3 on cv2(T2). Let T3 be the element of D that contains e3. By
Claim 2, T3 is of type D, which implies that there are two edges of M incident to
cv2(T2), a contradiction.
Now, consider the auxiliary directed graph DD in which V (DD) = D and
(T1, T2) ∈ A(DD) if and only if tr(T2) = cvi(T1) for some i ∈ {1, 2} (we ignore
possible multiple edges). It is clear that the elements of type A in D are partitioned into A-chains if and only if DDconsists of vertex-disjoint directed cycles and isolated
vertices. Note that if a vertex u is a (primary or secondary) connection vertex of an element T ∈ D, then T is an element of type A and u is incident to an edge in M ∩ E(T ). Therefore, every vertex of G is a connection vertex of at most one element of D, and hence, by Claim 3, every vertex of DD has in-degree precisely 1.
Note also that given two elements T1 and T2 we have tr(T1) 6= tr(T2), otherwise
there would be a vertex with two green in edges. This implies that every vertex of DD has out-degree at most 2. Now, if T1 and T2 are two elements of type A in D
such that cv1(T1) = tr(T2) = u, by Lemma 4, we have aux(T2) = cv2(T1), which
means that E(T1) ∪ E(T2)contains the ve edges in E(G) incident to u, and hence,
no other element of D contains edges incident to u. This implies that every vertex of DD has out-degree at most 1, and hence DD consists of vertex-disjoint directed
3.3 Admissible decompositions
In this section, we present a new decomposition invariant, which we call admis-sible decompositions, and conclude our proof. For that, we introduce an important object, the exceptional pair. Let G be a {g, r}-graph, and let D be a decomposition of G into trails of length 5. We say that a pair (T1, T2)of elements of D is an
excep-tional pair if T1and T2are elements of type A and C, respectively, and can be written
as T1 = a0a1a2a3a4a5 and T2 = b0b1b2b3b4b5 such that a2a3 ∈ Mg,r, a2 = a5 = b3, and
a2a1, a3a4, b2b1, b4b3 ∈ Fg, a4a5, b3b2, b4b5 ∈ Fr, a1a0, b1b0 ∈ Mg,r∪ Fg∪ Fr (see Figure
3.11). Note that since G is a simple graph, we have b4 6= a3. Also, if 2g + 2r 6= 0,
then we have b1 6= a3. This yields the following remark.
Remark 1. If G is a {g, r}-graph for which 2g+2r 6= 0 and (T1, T2)is an exceptional
pair, then T2 does not contain a hanging edge at cv(T1).
a0 a1 a2 a3 a4 b0 b1 b2 b4 b5
Figure 3.11: An exceptional pair.
An open chain is a sequence T0, T1, . . . , Ts−1 such that the following hold. (i) T0
is a free element of type A; (ii) Tj is an element of type A and tr(Tj) = cvi(Tj−1),
j ∈ {0, . . . , s − 2}and some i ∈ {1, 2}; and (iii) Ts−1 is an element of type C such
that (Ts−2, Ts−1) is an exceptional pair. The next step of our proof requires the
following variation of De nition 2.
De nition 3. We say that a decomposition D of a {g, r}-graph G into trails of length 5 is admissible if the following hold.
(a) There is a hanging edge at every connection vertex, except possibly for at most one secondary connection vertex cv2(T0), and, in this case, there is an open
chain S = T0, . . . , Ts−2, Ts−1 in D, such that Ts−2 = T0;
(b) Every element in D is either a path or an element of type A;
(c) If T ∈ D can be written as a0a1a2a3a4a5 and a0a1 ∈ M, then a0 = a1+ g + r;
(d) The elements of type A in D are partitioned into A-chains and at most one open chain.
It is not hard to check that the decomposition given by Lemma 5 is an admissible decomposition. Therefore, every {g, r}-graph such that 2g + 2r 6= 0 admits an admissible decomposition. By performing some more exchanging of edges between the elements of same A-chain of an admissible decomposition, we can show that an admissible decomposition that minimizes its number of elements of type A is in fact a P5-decomposition.
Theorem 3. Every {g, r}-graph for which 2g + 2r 6= 0 admits a P5-decomposition.
Proof. Let g and r be as in the statement, let G be a {g, r}-graph for which 2g+2r 6= 0, and put M = Mg,r. By Lemma 5, G admits an admissible decomposition. Let D
be an admissible decomposition of G that minimizes τ(D). In what follows, we prove that τ(D) = 0. Suppose, for a contradiction, that τ(D) > 0. We divide that A-chains into three types, according to the connections between its elements. Given i ∈ {1, 2}, we say that an A-chain S = T0, T1, . . . , Ts−1 is of type i if tr(Tj) = cvi(Tj−1)for every
j ∈ {0, . . . , s − 1}; and we say that S is a mixed A-chain if S is not of type 1 or 2. Analogously to the proof of Lemma 5, in each step, we exchange edges between some elements of D and obtain an admissible decomposition D0 into trails of length
ve such that τ(D0) < τ (D), which is a contradiction to the minimality of D. To
check that D0 is an admissible decomposition, we observe the four following items:
(i) The only vertex that has a hanging edge of D but does not have a hanging edge of D0 is the secondary connection vertex of an element T
1 of type A, and in this case
there is an element T2 of type C such that (T1, T2)is an exceptional pair, and hence
De nition 3(a) holds for D0; (ii) every element of D0 that is not an element of D, i.e.,
the elements involved in the exchange of edges, is a path or an element of type A, and hence 3(b) holds for D0; (iii) every edge uv ∈ M in the elements considered that
can be viewed as an end edge in D0 (some elements may be expressed by more than
two trails) either can be viewed as end edges in D or is obtained from an element of type A and satis es u = v + g + r, and hence De nition 3(c) holds for D0; (iv)
either an open chain is reduced by one element, an A-chain is converted into an open chain, or all the elements of an A-chain are replaced by paths of length 5. Claim 4. Every A-chain in D is mixed.
Proof. Suppose, for a contradiction, that there is an A-chain S = T0, T1, . . . , Ts−1
of type 1 or 2. Let Tj = a0,ja1,ja2,ja3,ja4,ja5,j, where a5,j = a2,j, a2,ja3,j ∈ M,
a1,ja0,j, a3,ja4,j ∈ Fg, a4,ja2,j ∈ Fr and a2,ja1,j ∈ M ∪ Fg∪ Fr. For i ∈ {1, 2, 3, 4, 5},
the edge ai−1,jai,j is called the i-th edge of Tj. In what follows, we divide the proof
according to the type of S.
S is of type 1. Since S is of type 1, for each j = 0, . . . , s − 1, we have a3,j =
aux(Tj+1) = a1,j+1. Now, for each j = 0, . . . , s−1, let Tj0 = a2,j+1a3,ja4,ja1,ja2,ja0,j+1
(see Figure 3.12). Note that T0
j = Tj− a1,ja0,j+ a1,j+a0,j+1− a2,ja3,j+ a2,j−1a3,j−1−
a4,ja2,j + a4,j+1a2,j+1. More speci cally, we have a2,j+1a3,j = a4,j+1a5,j+1 is the 5-th
edge of Tj+1; a3,ja4,j is the 4-th edge of Tj; a4,ja1,j = a2,j−1a3,j−1 is the 3rd edge of
Tj−1; a1,ja2,j is the 2nd edge of Tj; a2,ja0,j+1 = a1,j+1a0,j+1 is the 1st edge of Tj+1.
Clearly, T0
j is a trail of length 5. Moreover, since, for each i ∈ {1, 2, 3, 4, 5}, the
element T0
j contains the i-th edge of an element of S, and, if j 6= j
0, the elements T0 j
and T0
j0 contain the i-th edge of dierent elements of S, the set D0 = D \ {Tj: j =
0, . . . , s − 1} ∪ {T0
j: j = 0, . . . , s − 1} is a decomposition of G into trails of length 5.
We may regard D0 as the decomposition obtained by reversing the direction of two
components of Fg, namely, the green edges in S, and applying the same strategy
used in Proposition 3.
In order to prove that T0
j is a path, we show that a2,j+1, a0,j+1 ∈/
{a3,j, a4,j, a1,j, a2,j}. Note that, since, for each j ∈ {0, . . . , s − 1}, Tj is a path,
we have ai,j 6= ai0,j for every i 6= i0. Since G is a simple graph, we have
a2,j+1 ∈ {a/ 3,j, a4,j, a2,j} and a0,j+1 ∈ {a/ 3,j, a4,j, a1,j, a2,j}; and if a2,j+1 = a1,j, then
a4,j+1a2,j+1 and a4,j−1,a2,j−1 are two distinct red in-edges of a1,j, a contradiction.
a2,j+1 a3,j a4,j a1,j a2,j a0,j+1
Figure 3.12: Exchange of edges between the elements an A-chain of type 1 with ve elements in the proof of Claim 4.
We claim that D0 is an admissible decomposition. Indeed, the only vertices of
the elements of S that can be connection vertices of elements in D0 are the vertices
a0,j, for j = 0, . . . , s−1. But a hanging edge at a vertex a0,j is in Tj0 ∈ Dif and only
if a0,j = a3,j0 for some j0 6= j, and, in this case a3,j0 is not a connection vertex in D0
because all edges incident to it are in elements of {T0
j: j = 0, . . . , s − 1}. Therefore,
De nition 3(a) holds for D0. Moreover, since T0
j is a path, for j = 0, . . . , s − 1,
De nition 3(b) holds for D0. Also, every edge of M in an element of S is the
middle edge of an element T0
j, for some j ∈ {0, . . . , s − 1}, and hence De nition 3(c)
holds for D0. Finally, D and D0 have the same number of open chain, and hence
that τ(D0) = τ (D) − s, a contradiction to the minimality of D.
Sis of type 2. Since S is of type 2, for each j = 0, . . . , s−1, we have a2,j = cv2(Tj) =
tr(Tj+1) = a4,j+1. Now, for each j = 0, . . . , s − 1, let Tj0 = a0,ja1,ja2,ja3,ja4,ja4,j−1
(see Figure 3.13). Clearly, T0
j is a trail of length 5. Note that Tj0 = Tj − a4,ja2,j +
a4,j−1a2,j−1, i.e., Tj0 is the element obtained from Tj by exchanging its 5-th edge by
the 5-th edge of Tj−1. Thus, the set D0 = D \ {Tj: j = 0, . . . , s − 1} ∪ {Tj0: j =
0, . . . , s − 1} is a decomposition of G into trails of length 5. We may regard D0 as the decomposition obtained by reversing the direction of one component of Fr
and applying the same strategy used in Proposition 3. In order to prove that T0 j is
a path, we show that a4,j−1 ∈ {a/ 0,j, a1,j, a2,j, a3,j, a4,j}. Note that, since, for each
j ∈ {0, . . . , s−1}, Tj is a path, we have ai,j 6= ai0,jfor every i 6= i0. Since G is a simple
graph, we have a4,j−1 ∈ {a/ 2,j, a3,j, a4,j}; also, by Lemma 2, we have a4,j−1 6= a0,j;
and if a4,j−1 = a1,j, then a2,ja1,j and a3,j−1a4,j−1 are two distinct green in-edges of
a4,j−1, a contradiction. a0,j a1,j a2,j a4,j a4,j−1 a3,j
Figure 3.13: Exchange of edges between the elements an A-chain of type 2 with ve elements in the proof of Claim 4.
We claim that D0 is an admissible decomposition. Indeed, the only vertices that
have hanging edges in D and may not have hanging edges in D0 are the vertices a 3,j
and a4,j = a2,j−1, for j = 0, . . . , s−1, but these vertices are connection vertices of the
elements in S, and hence can't be connection vertices of elements in D0. Therefore,
De nition 3(a) holds for D0. Moreover, since T0
j is a path, for j = 0, . . . , s − 1,
De nition 3(b) holds for D0. Also, every edge of M in an element of S is either the
middle edge or an end edge of an element T0
j, for some j ∈ {0, . . . , s − 1}, and hence
De nition 3(c) holds for D0. Finally, D and D0 have the same number of open chain,
and hence De nition 3(d) holds for D0. Therefore, D0is an admissible decomposition
of G such that τ(D0) = τ (D) − s, a contradiction to the minimality of D.
Claim 5. Every A-chain contains at least four elements
Proof. Let S be an A-chain in D with at most three elements. First, note that if S contains two elements, then G contains a parallel edge, a contradiction. Therefore,
we may assume that S contains three elements T1, T2 and T3. By Claim 4, we may
assume tr(T1) = cv2(T3), tr(T2) = cv1(T1) and tr(T3) = cvi(T2), for i ∈ {1, 2}. In
what follows, we divide the proof depending on whether i = 1 or i = 2.
Let T1 = a0a1a2a3a4a5, T2 = b0b1b2b3b4b5 and T3 = c0c1c2c3c4c5 be the elements
of S where a4 = tr(T1) = cv2(T3) = c2, b4 = tr(T2) = cv1(T1) = a3 and c4 = tr(T3) =
cvi(T2).
Case i = 1. In this case, c4 = tr(T3) = cv1(T2) = b3. Since b3 = c4, we have
b2 = c1 and c0 = a1. Put T10 = a0a1a2a3a4c1, T20 = b0b1b2b4b3c2, T30 = c0c1c4c3c2a2 and D0 = D \ {T 1, T2, T3} ∪ {T10, T 0 2, T 0 3}. Note that D 0 is a decomposition of G into
trails of length 5. We claim that T0 1, T
0
2 and T 0
3 are all paths. By Lemma 3, T 0 2 is
a path. Since G is simple, c1 ∈ {a/ 2, a3, a4} and a2 ∈ {c/ 1, c2, c3, c4}. By Lemma 2,
we have c1 6= a0 and a2 6= c0. Therefore, T30 is of type a path. Finally, if c1 = a1,
then c2c1 and a2a1 are two green out-edges at a1, a contradiction. Therefore, T10 is
a path.
To check that D0is an admissible decomposition rst note that hang
D0(cvi(Tj)) ≥
hangD(cvi(Tj)) for every trail Tj ∈ D0. Thus, de nition 3(a) holds for D. As
seen above, the new elements are all paths. Thus, de nition 3(b) holds. Since D is admissible and the new elements are paths, the elements of type A are still partitioned into A-chains. Thus, 3(d) holds for D0. Also, since T
1 is an element of
type A, we have a2 = a3+ g + r, and hence De nition 3(c) holds for D0. Therefore,
D0 is an admissible decomposition of G such that τ(D0) = τ (D) − 3, a contradiction
to the minimality of D. a0 a1 a3= b4 a4 b0 b1= a2 b3= c4 c0 c1= b2 c2 c3 a0 a1 a3= b4 a4 b0 b1= a2 b3= c4 c0 c1= b2 c2 c3
Figure 3.14: Exchange performed in the proof of Claim 5 in the case tr(T3) = cv1(T2).
Case i = 2. In this case, c4 = tr(T3) = cv2(T2) = b2. Put T10 = a0a1a2a4a3b5,
T20 = b0b1b4b3b2c2, T30 = c0c1c2c3c4b1 and let D0 = D \ {T1, T2, T3} ∪ {T10, T 0 2, T
0 3}.
Note that D0 is a decomposition of G into trails of length 5. In what follows, we prove
that T0
1, T20 and T30 are all paths. By Lemma 3, T10 is a path. Since G is simple, we
have c2 ∈ {b/ 1, b2, b3, b4}and b1 ∈ {c/ 2, c3, c4}. By Lemma 2, b2 6= a0, c2 6= b0, b1 6= c0.
Therefore, T0
2 is a path. If b1 = c1, then b2b1 and c2c1 are two green in-edges at c1,
a contradiction. Therefore, T0
3 is also a path.
To check that D0is an admissible decomposition rst note that hang
hangD(cvi(Tj)) for every trail Tj ∈ D0. Thus, de nition 3(a) holds for D. As
seen above, the new elements are all paths. Thus, de nition 3(b) holds. Since D is admissible and the new elements are paths, the elements of type A are still partitioned into A-chains. Thus, 3(d) holds for D0. Also, since T
1 and T3 are
elements of type A, we have a2 = a3+ g + r, and hence De nition 3(c) holds for D0.
Therefore, D0 is an admissible decomposition of G such that τ(D0) = τ (D) − 3, a
contradiction to the minimality of D.
a0 a1 a3= b4 a4 b0= c1 b1= a2 b2= c4 b3 c0 c2 c3 a0 a1 a3= b4 a4 b0= c1 b1= a2 b2= c4 b3 c0 c2 c3
Figure 3.15: Exchange performed in the proof of Claim 5 in the case tr(T3) = cv2(T2)
Claim 6. D contains an open chain.
Proof. Suppose, for a contradiction, that there is no open chain in D. Since τ(D) > 0, D contains an A-chain S = T0, T1, . . . , Ts−1. By Claim 4, S is a mixed A-chain.
Then we can nd three consecutive elements in S, say Tj, Tj+1, Tj+2, such that
cv2(Tj) = tr(Tj+1)and cv(Tj+1) = tr(Tj+2). Since S is cyclic, i.e., tr(T0) = cvi(Ts−1),
for some i ∈ {1, 2}, we may suppose, without loss of generality, that j = 0. By Claim 5, there is an element T3 ∈ D such that tr(T3) = cvi(T2), for some i ∈ {1, 2}.
In what follows, the proof is divided according to i. Let T0 = a0a1a2a3a4a5, T1 =
b0b1b2b3b4b5, T2 = c0c1c2c3c4c5, and T3 = d0d1d2d3d4d5, where a5 = a2, b5 = b2,
c5 = c2, d5 = d2, and a2a3, b2b3, c2c3, d2d3 ∈ M. By the choice of T0, T1, and T2,
we have b4 = tr(T1) = cv2(T0) = a2, c4 = tr(T2) = cv1(T1) = b3. The exchanges
of edges performed here are analogous to the exchanges performed on the proof of Claim 2 of Lemma 5 for elements of type A.
Case tr(T3) = cv1(T2). In this case, we have d4 = c3 and, by Lemma 4, d2 = c1.
Put T0
1 = b0b1b2b4b3c2, T20 = c0c1c4c3c2d0, T30 = c1d1d2d3d4d2 (see Figure 3.16), and
let D0 = D \ {T
1, T2, T3} ∪ {T10, T20, T30}. Note that D0 is a decomposition of G
into trails of length 5. By Lemma 3, T0
1 is an element of type C. In what follows,
we prove that T0
2 is a path and T 0
3 is an element of type A. Since G is simple, we
have d0 ∈ {c/ 1, c2, c3, c4}, and c1 ∈ {d/ 1, d2, d4}. By Lemma 2, d0 6= c0, and hence,
T20 is a path. If c1 = d3, then b2b3 and d2d3 are two edges of M incident to c1, a
contradiction. Therefore, T0
b0 b1 b3= c4 b4 c0 c1= b2 c3= d4 d0 d1= c2 d2 d3 b0 b1 b3= c4 b4 c0 c1= b2 c3= d4 d0 d1= c2 d2 d3
Figure 3.16: Exchange performed in the proof of Claim 6 in the case tr(T3) = cv1(T2)
.
To check that D0 is an admissible decomposition rst note that hang
D0(v) ≥
hangD(v) for every v ∈ V (G) \ {a2}. Note also that b4b2 is a hanging edge at a2 in
D but not in D0. However, (T
1, T20) is an exceptional pair. Thus, since c3 is not a
connection vertex in D0, the element T0
3 is free. Therefore, S 0 = T0
3, . . . Ts−1T1T20 is
an open chain, and hence De nition 3(a) holds for D0. Since, T0
1 and T20 are paths
and T0
3 is an element of type A, De nition 3(b) holds for D0. Since no edge of M
that is not a hanging edge in D is a hanging edge in D0, De nition 3(c) holds for
D0. Finally, note that an element T of type A in D \ {T
1, T2, T3} is either in an
A-chain of D dierent from S, which implies that T is in an A-chain of D0, or is in
S, which implies that T is in S0. Thus, De nition 3(d) holds for D0. Therefore, D0 is an admissible decomposition of G such that τ(D0) = τ (D) − 2, a contradiction to
the minimality of D.
Case tr(T3) = cv2(T2). In this case, we have d4 = c2. Put T10 = b0b1b2b4b3c2, T20 =
c0c1c4c3c2d2, T30 = c1d4d3d2d1d0 (see Figure 3.17), and let D0 = D \ {T1, T2, T3} ∪
{T0 1, T 0 2, T 0 3}. Note that D
0 is a decomposition of G into trails of length 5. By
Lemma 3, T0
1 is an element of type C. In what follows, we prove that T 0
2 and T 0 3
are paths. Since G is simple, we have d2 ∈ {c/ 1, c2, c3, c4}, and c1 ∈ {d/ 2, d3, d4}. By
Lemma 2, d2 6= c0, and c1 6= d0. Therefore, T20 is a path. If c1 = d1, then d2d1 and
c2c1 are two green in edges of c1. Therefore, T30 is a path.
b0 b1 b3= c4 b4 c0= d1 c1= b2 c2= d4 c3 d0 d2 d3 b0 b1 b3= c4 b4 c0= d1 c1= b2 c2= d4 c3 d0 d2 d3
Figure 3.17: Exchange performed in the proof of Claim 6 in the case tr(T3) = cv2(T2)
.
To check that D0 is an admissible decomposition rst note that hang
D0(v) ≥
D but not in D0. However, (T
1, T20) is an exceptional pair. Thus, since d2 and d3
are not connection vertices in D0, the element T
4 (or T1, if s = 4) is free. Therefore,
S0 = T4, . . . Ts−1T1T20 is an open chain, and hence De nition 3(a) holds for D0. Since,
T10, T20 and T30 are paths, De nition 3(b) holds for D0. Since no edge of M that is not a hanging edge in D is a hanging edge in D0, De nition 3(c) holds for D0. Finally,
note that an element T of type A in D \ {T1, T2, T3} is either in an A-chain of D
dierent from S, which implies that T is in an A-chain of D0, or is in S, which implies
that T is in S0. Thus, De nition 3(d) holds for D0. Therefore, D0 is an admissible
decomposition of G such that τ(D0) = τ (D) − 3, a contradiction to the minimality
of D.
Let S = T0, T1, . . . , Ts−1 be an open chain in D. Let Tj = a0,ja1,ja2,ja3,ja4,ja5,j,
for j ∈ {0, . . . , s − 1}, where a2,ja3,j ∈ M and a5,j = a2,j for j ∈ {0, . . . , s − 2}.
Claim 7. T1 is an element of type A and tr(T1) = cv1(T0).
Proof. Suppose, for a contradiction, that T1 is not an element of type A or tr(T1) =
cv2(T0). We claim that T1 does not contain a hanging edge at cv1(T0). Indeed,
if T1 is not an element of type A, then, by the de nition of open chain, T1 is an
element of type C, and hence, by Remark 1, T1 does not have a hanging edge
at cv1(T0); and if T1 is an element of type A for which tr(T1) = cv2(T0), then
we have a4,1 = tr(T1) = cv2(T0) = a2,0, and hence, if a1,1 = a3,0, then we have
a4,1 + r + g = a1,1 = a3,0 = a2,0 − r − g, which implies that 2g + 2r = 0, a
contradiction. Therefore, T1 does not contain a hanging edge at cv1(T0).
However, by De nition 3(a), there is a hanging edge at cv1(T0). Thus, let T =
u0u1u2u3u4u5 be an element of D \ {T0, T1} that contains a hanging edge, say u1u0,
at cv1(T0). Note that all the edges incident to a2,0 are in E(T0) ∪ E(T1). Let T00 =
a0,0a1,0a2,0a4,0a3,0u0 and T0 = a2,0u1u2u3u4u5 and put D0 = D \ {T0, T0} ∪ {T00, T 0}.
By Lemma 3, T0
0is a path; and since all the edges incident to a2,0are in E(T0)∪E(T1),
we have a2,0 ∈ {u/ 1, u2, u3, u4, u5}, and hence if T is a path (resp. an element of type
A), then T0 is a path (resp. an element of type A). Thus De nition 3(b) holds for
D0.
To check that D0 is an admissible decomposition rst note that hang
D0(v) ≥
hangD(v) for every v ∈ V (G) \ {a4,0}. Thus, since T0 is a free element, a4,0 is not
a connection vertex in D, and hence a4,0 is not a connection vertex in D0. Note
also that T1 is either an element of type C or a free element of type A, and hence
S0 = T1, . . . , Ts−1 is an open chain. Thus, De nition 3(a) holds for D0. Also, since
T0 is an element of type A, we have a2,0 = a3,0 + g + r, and hence De nition 3(c)
holds for D0. Finally, note that an element T of type A in D \ {T
0} is either in an
A-chain of D dierent from S, which implies that T is in an A-chain of D0, or is in
is an admissible decomposition of G such that τ(D0) = τ (D) − 1, a contradiction to
the minimality of D.
By Claim 7, we have s ≥ 3, and hence, there is an element T2 in S. Note that,
by Lemma 4, since tr(T1) = cv1(T0), we have aux(T1) = cv2(T0). This implies that
a1,1a0,1 ∈ Fr because all the edges incident to a1,1 = cv2(T0) are in E(T0) ∪ E(T1).
Claim 8. T2 is of type A.
Proof. Suppose, for a contradiction, that T2 is not of type A, then T2 is an
element of type C and (T1, T2) is an exceptional pair. Thus, we can write
T2 = a0,2a1,2a2,2a3,2a4,2a5,2 such that a2,1 = a5,1 = a3,2, and a2,2a1,2, a4,2a3,2 ∈ Fg,
a3,2a2,2, a4,2a5,2 ∈ Fr, a1,2a0,2 ∈ Mg,r ∪ Fg ∪ Fr. We claim that a0,1 = a1,2. Indeed,
since a1,1a0,1 ∈ Fr, we have a0,1 = a1,1 + r = a2,1 + g + r, but by the de nition
of type C, we have a1,2 = a2,2 + g = a3,2+ r + g. Thus, since a3,2 = a2,1, we
ob-tain a0,1 = a1,2. Now, put T00 = a0,0a1,0a2,0a4,0a3,0a2,1, T10 = a1,1a4,1a3,1a2,1a2,2a0,1,
and T0 2 = a0,2a1,2a1,1a3,2a4,2a5,2, and put D0 = D \ {T0, T1, T2} ∪ {T00, T 0 1, T 0 2}. By Lemma 3, T0
0 is a path; since G is a simple graph, a2,2 ∈ {a/ 1,1, a4,1, a3,1, a2,1, a2,2, a0,1},
and hence T0
1 is a path; and since all edges incident to a1,1 are in E(T0) ∪ E(T1), we
have a1,1 ∈ V (T/ 2), which implies that T20 is a path. Thus De nition 3(b) holds for
D0. a0,0 a1,0 a2,0= a1,1 a3,0= a4,1 a4,0 a0,1= a1,2 a2,1= a3,2 a3,1 a0,2 a2,2 a4,2 a5,2 a0,0 a1,0 a2,0= a1,1 a3,0= a4,1 a4,0 a0,1= a1,2 a2,1= a3,2 a3,1 a0,2 a2,2 a4,2 a5,2
Figure 3.18: Exchange of edges between two elements of type A and another element of type E in the proof of Claim 8.
To check that D0 is an admissible decomposition rst note that hang
D0(v) ≥
hangD(v)for every v ∈ V (G) \ {a4,0, a1,1, a3,1}. Thus, since T0 is a free element, a4,0
is not a connection vertex in D, and hence a4,0 is not a connection vertex in D0;
and since the edges of M incident to a1,1 and a3,1 are in T10, a1,1 and a3,1 are not
connection vertices in D0. Note also that no element of S is in D0, and hence there
are no open chains in D0. Thus, De nitions 3(a) and 3(d) hold for D0. Also, since T 0
is an element of type A, we have a2,0 = a3,0+g+r, and hence De nition 3(c) holds for
D0. Therefore, D0 is an admissible decomposition of G such that τ(D0) = τ (D) − 2,
Now, by Claim 8, we have s ≥ 4. In what follows, we divide the proof depending on whether tr(T2) = cv1(T1) or tr(T2) = cv2(T1). This proof is analogous to the
proof of Claim 6.
Case tr(T2) = cv1(T1). By Lemma 4, we have a1,2 = aux(T2) = cv2(T1) = a2,1. Put
T00 = a0,0a1,0a2,0a4,0a3,0a2,1, T10 = a0,1a1,1a4,1a3,1a2,1a0,2, T20 = a1,1a1,2a2,2a3,2a4,2a2,2
(see Figure 3.19), let D0 = D \ {T
0, T1, T2} ∪ {T00, T10, T20}, and let S0 =
T20, T3, . . . , Ts−1. Note that D0 is a decomposition of G into trails of length 5. By
Lemma 3, T0
0 is an element of type C. In what follows, we prove that T 0
1is a path and
T20 is an element of type A. Since G is simple, we have a0,2 ∈ {a/ 1,1, a2,1, a3,1, a4,1},
and a1,1 ∈ {a/ 1,2, a2,2, a4,2}. By Lemma 2, a0,2 6= a0,1, and hence, T10 is a path. If
a1,1 = a3,2, then a2,0a3,0 and a2,2a3,2 are two edges of M incident to a1,1, a
contra-diction. Therefore, T0 2 is an element of type A. a0,0 a1,0 a3,0= a4,1 a4,0 a0,1 a1,1= a2,0 a3,1= a4,2 a0,2 a1,2= a2,1 a2,2 a3,2 a0,0 a1,0 a3,0= a4,1 a4,0 a0,1 a1,1= a2,0 a3,1= a4,2 a0,2 a1,2= a2,1 a2,2 a3,2
Figure 3.19: Exchange performed in the proof of Claim 6 in the case tr(T3) = cv1(T2)
.
To check that D0 is an admissible decomposition rst note that hang
D0(v) ≥
hangD(v) for every v ∈ V (G) \ {a4,0}, but since T0 is free, a4,0 is not a connection
vertex in D, and hence a4,0 is not a connection vertex in D0. Thus, since a3,1 is not a
connection vertex in D0, the element T0
2 is free. Therefore, S0 is an open chain, and
hence De nition 3(a) holds for D0. Since, T0
0 and T10 are paths and T20 is an element
of type A, De nition 3(b) holds for D0. Since no edge of M that is not a hanging
edge in D is a hanging edge in D0, De nition 3(c) holds for D0. Finally, note that an
element T of type A in D \ {T0, T1, T2}is either in an A-chain of D dierent from S,
which implies that T is in an A-chain of D0, or is in S, which implies that T is in S0.
Thus, De nition 3(d) holds for D0. Therefore, D0 is an admissible decomposition of
Gsuch that τ(D0) = τ (D) − 2, a contradiction to the minimality of D.
Case tr(T3) = cv2(T2). Put T00 = a0,0a1,0a2,0a4,0a3,0a2,1, T10 = a0,1a1,1a4,1a3,1a2,1a2,2,
T20 = a0,2a1,2a2,2a3,2a4,2a1,1(see Figure 3.20), let D0 = D \{T0, T1, T2} ∪{T00, T 0 1, T
0 2},
and let S0 = T
3, . . . , Ts−1. Note that D0 is a decomposition of G into trails of
length 5. By Lemma 3, T0
1 is an element of type C. In what follows, we prove that
T10 and T20 are paths. Since G is simple, we have a2,2 ∈ {a/ 1,1, a2,1, a3,1, a4,1}, and
